Heavy Bikes are better !
#126
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The only dilemma here if riding the same route and comparing between heavy and light bike, you'll arrive at the destination sooner with the lighter bike if spending the same effort. That means less calories burned because you were riding at the same effort you would with a heavier bike but in a shorter period of time.
To actually achieve the same workout level with a heavier bike with a lighter bike, you'll have to pedal at a higher effort than you would with a heavier bike. Of course, this is only feasible to do if it's safe (to cruise at significantly higher speeds)
To actually achieve the same workout level with a heavier bike with a lighter bike, you'll have to pedal at a higher effort than you would with a heavier bike. Of course, this is only feasible to do if it's safe (to cruise at significantly higher speeds)
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#127
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I don't know if I've ever ridden into a headwind that strong, but my experience with headwinds generally is you push hard between the gusts and/or you adjust your direction so it's not fully in your face. There is a point where I find the strength of crosswinds make it impossible to safely operate the bike, but I don't know what that is in KPH.
In actual wind, I do try push through between the gusts.
On flat ground, I can hit around 40-45 kph before wishing for a heavier gear (not that good at high rpm spinning yet - I'm a natural masher), but truth be told, I'm not sure I could push the bike much faster on flat ground in a much heavier gear (at least this particular bike).
If you double your speed, you quadruple the wind resistance (force).
So, going 9 mph to 18 mph results in four times the resistance, and going from 18 mph to 36mph is four times that - which means that 36mph has 16 times the wind resistance as 9 mph. Yes, I did choose 9 mph because it came close to 37mph when multiplied by four, lol.
This is the reason for those time trial setups and why no one but Cubewheels stand up in strong headwinds.
60 kph/37mph is a Beaufort 7 - a "near gale". Something you most likely have experienced quite a few times, as most of us on here have (because it's not a rare wind strength).
Edited to add:
If going 20 kph (12.5 mph) into a 60kph/37mph headwind, the effective wind is 50 mph. That's almost highway speeds. There's an immense wind resistance at those wind speeds.
Last edited by CargoDane; 10-30-20 at 07:32 AM.
#128
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60 kph is 37 mph. For comparison, When I hit 45 kph (downhill) I can really feel the wind. At 55 kph I spin out with my gearing (I am trying to become better at spinning).
In actual wind, I do try push through between the gusts.
On flat ground, I can hit around 40-45 kph before wishing for a heavier gear (not that good at high rpm spinning yet - I'm a natural masher), but truth be told, I'm not sure I could push the bike much faster on flat ground in a much heavier gear (at least this particular bike).
If you double your speed, you quadruple the wind resistance (force).
So, going 9 mph to 18 mph results in four times the resistance, and going from 18 mph to 36mph is four times that - which means that 36mph has 16 times the wind resistance as 9 mph. Yes, I did choose 9 mph because it came close to 37mph when multiplied by four, lol.
This is the reason for those time trial setups and why no one but Cubewheels stand up in strong headwinds.
60 kph/37mph is a Beaufort 7 - a "near gale". Something you most likely have experienced quite a few times, as most of us on here have (because it's not a rare wind strength).
Edited to add:
If going 20 kph (12.5 mph) into a 60kph/37mph headwind, the effective wind is 50 mph. That's almost highway speeds. There's an immense wind resistance at those wind speeds.
In actual wind, I do try push through between the gusts.
On flat ground, I can hit around 40-45 kph before wishing for a heavier gear (not that good at high rpm spinning yet - I'm a natural masher), but truth be told, I'm not sure I could push the bike much faster on flat ground in a much heavier gear (at least this particular bike).
If you double your speed, you quadruple the wind resistance (force).
So, going 9 mph to 18 mph results in four times the resistance, and going from 18 mph to 36mph is four times that - which means that 36mph has 16 times the wind resistance as 9 mph. Yes, I did choose 9 mph because it came close to 37mph when multiplied by four, lol.
This is the reason for those time trial setups and why no one but Cubewheels stand up in strong headwinds.
60 kph/37mph is a Beaufort 7 - a "near gale". Something you most likely have experienced quite a few times, as most of us on here have (because it's not a rare wind strength).
Edited to add:
If going 20 kph (12.5 mph) into a 60kph/37mph headwind, the effective wind is 50 mph. That's almost highway speeds. There's an immense wind resistance at those wind speeds.
Like I said above, I am not defending his wind physics because I don't believe they're correct, just explaining how I deal with high headwinds in a manner that mitigates the aero effects. I'm not disagreeing with your math, I'm saying that people who are riding in such winds aren't really making progress except between gusts or when they can aim the bike out of the direct blast. Remember that a strong headwind can also be one hell of a great tailwind, and there's a lot of on-between impacts depending on where it's hitting you.
You and I aren't really disagreeing at all.
#129
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Like I said above, I am not defending his wind physics because I don't believe they're correct, just explaining how I deal with high headwinds in a manner that mitigates the aero effects. I'm not disagreeing with your math, I'm saying that people who are riding in such winds aren't really making progress except between gusts or when they can aim the bike out of the direct blast. Remember that a strong headwind can also be one hell of a great tailwind, and there's a lot of on-between impacts depending on where it's hitting you.
You and I aren't really disagreeing at all.
You and I aren't really disagreeing at all.
#130
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Now we're all in agreement we'd be meeting the same forces in terms of wind drag forces...
The formula to get power required to overcome that force is Power = Force x Velocity. That formulas shows us it takes only 1/4 the power to cruise 20 kph with 60 kph headwind vs cruising at 80 kph no wind
Formula found here (first box - the simplified formula):
Power
The formula to get power required to overcome that force is Power = Force x Velocity. That formulas shows us it takes only 1/4 the power to cruise 20 kph with 60 kph headwind vs cruising at 80 kph no wind
Formula found here (first box - the simplified formula):
Power
The main force we are worried are the wind force. So it's the effective velocity of the wind. Not the velocity of the bike as the only thing thing that has an influence here on the actual bike is the rolling resistance. To find the wind resistance, all you need to concern yourself with is the effective wind velocity.
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Now we're all in agreement we'd be meeting the same forces in terms of wind drag forces...
The formula to get power required to overcome that force is Power = Force x Velocity. That formulas shows us it takes only 1/4 the power to cruise 20 kph with 60 kph headwind vs cruising at 80 kph no wind
Formula found here (first box - the simplified formula):
Power
The formula to get power required to overcome that force is Power = Force x Velocity. That formulas shows us it takes only 1/4 the power to cruise 20 kph with 60 kph headwind vs cruising at 80 kph no wind
Formula found here (first box - the simplified formula):
Power
The air drag component of cycling work is k(a)*A*v**2*s, where s is the bicycle speed on the road, k(a) is the air resistance coefficient, A is the combined frontal area of cyclist and bicycle, and v is the bicycle speed through the air (i.e. road speed plus head wind speed).
So, the v**2 term is the same and the difference is the value of s: 20 vs 80.
Otto
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That link doesn’t work for me, but the result is correct.
The air drag component of cycling work is k(a)*A*v**2*s, where s is the bicycle speed on the road, k(a) is the air resistance coefficient, A is the combined frontal area of cyclist and bicycle, and v is the bicycle speed through the air (i.e. road speed plus head wind speed).
So, the v**2 term is the same and the difference is the value of s: 20 vs 80.
Otto
The air drag component of cycling work is k(a)*A*v**2*s, where s is the bicycle speed on the road, k(a) is the air resistance coefficient, A is the combined frontal area of cyclist and bicycle, and v is the bicycle speed through the air (i.e. road speed plus head wind speed).
So, the v**2 term is the same and the difference is the value of s: 20 vs 80.
Otto
To calculate the force of wind resistance, you cannot ignore actual wind speed in relation to the ground (and you). If you could, it would be equally hard to ride 20 kph in still air, in 60 kph headwinds, and in 60 kph tail winds as there is no effective wind resistance at all other than what results from your own progress over the ground.
Wind resistance (drag) is calculated against the speed of the wind, not the speed over ground. Hence why planes go slower over the ground in a headwind, and faster with a tail wind, yet they travel through the air at the same speed.
This is also the reason that wind tunnels work. If you calculated drag as speed over ground, wind tunnels would be absolutely useless.
To put it another way: Wind resistance (drag) is a measure of how much resistance there is for going through the air at a given speed.
Last edited by CargoDane; 10-30-20 at 08:53 AM. Reason: added a bit
#135
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I bet we could probably learn a lot about heavy bikes from those guys who drive bicycle taxis in various parts of the world. I am no specially skilled rider
but I have a tandem bike and I give rides to a number of people who are otherwise unable to peddle. i have a place in the frame where I ask them to
keep their feet so as to not interfere with the cranks going round and round. Hard to put into words but you do indeed gain a unique perspective.
I also ride my tandem solo which is unique.... I say it feels like a bicycle limousine. One thing I have to say i like a little bit is the fact that the front wheel steering is far less twitchy and might be more steady if I was riding hands off .... which I would never do.
but I have a tandem bike and I give rides to a number of people who are otherwise unable to peddle. i have a place in the frame where I ask them to
keep their feet so as to not interfere with the cranks going round and round. Hard to put into words but you do indeed gain a unique perspective.
I also ride my tandem solo which is unique.... I say it feels like a bicycle limousine. One thing I have to say i like a little bit is the fact that the front wheel steering is far less twitchy and might be more steady if I was riding hands off .... which I would never do.
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My bike isn't light. Nor heavy. By the time I load her up she feels heavy. A lighter bike would be faster. But she takes me where I want to go, mostly. You just have to figure out what you want.
This thread reminded me of an episode of "Insider Training" with Laird Hamilton. They showed him and somebody else loading up mountain bikes with weightlifting plates and riding dirt and gravel for the extra exercise. That looked heavy.
This thread reminded me of an episode of "Insider Training" with Laird Hamilton. They showed him and somebody else loading up mountain bikes with weightlifting plates and riding dirt and gravel for the extra exercise. That looked heavy.
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Again: That would be true if we weren't talking about wind, but merely talking about wind resistance in still air.
To calculate the force of wind resistance, you cannot ignore actual wind speed in relation to the ground (and you). If you could, it would be equally hard to ride 20 kph in still air, in 60 kph headwinds, and in 60 kph tail winds as there is no effective wind resistance at all other than what results from your own progress over the ground.
Wind resistance (drag) is calculated against the speed of the wind, not the speed over ground. Hence why planes go slower over the ground in a headwind, and faster with a tail wind, yet they travel through the air at the same speed.
This is also the reason that wind tunnels work. If you calculated drag as speed over ground, wind tunnels would be absolutely useless.
To put it another way: Wind resistance (drag) is a measure of how much resistance there is for going through the air at a given speed.
To calculate the force of wind resistance, you cannot ignore actual wind speed in relation to the ground (and you). If you could, it would be equally hard to ride 20 kph in still air, in 60 kph headwinds, and in 60 kph tail winds as there is no effective wind resistance at all other than what results from your own progress over the ground.
Wind resistance (drag) is calculated against the speed of the wind, not the speed over ground. Hence why planes go slower over the ground in a headwind, and faster with a tail wind, yet they travel through the air at the same speed.
This is also the reason that wind tunnels work. If you calculated drag as speed over ground, wind tunnels would be absolutely useless.
To put it another way: Wind resistance (drag) is a measure of how much resistance there is for going through the air at a given speed.
During WWII, the original mission for the B-29 bomber was high level bombing of Japan. However, it turns out that the jet stream flows directly over Japan at that altitude, so that the bombers were attempting to fly directly INTO the jet stream while dropping their load. The planes were often going about 5 mph in reverse while still maintaining lift. Needless to say, it was absolutely impossible to control where the bombs landed under those conditions, and the planes started flying their missions at a much lower altitude.
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Again: That would be true if we weren't talking about wind, but merely talking about wind resistance in still air.
To calculate the force of wind resistance, you cannot ignore actual wind speed in relation to the ground (and you). If you could, it would be equally hard to ride 20 kph in still air, in 60 kph headwinds, and in 60 kph tail winds as there is no effective wind resistance at all other than what results from your own progress over the ground.
Wind resistance (drag) is calculated against the speed of the wind, not the speed over ground. Hence why planes go slower over the ground in a headwind, and faster with a tail wind, yet they travel through the air at the same speed.
This is also the reason that wind tunnels work. If you calculated drag as speed over ground, wind tunnels would be absolutely useless.
To put it another way: Wind resistance (drag) is a measure of how much resistance there is for going through the air at a given speed.
To calculate the force of wind resistance, you cannot ignore actual wind speed in relation to the ground (and you). If you could, it would be equally hard to ride 20 kph in still air, in 60 kph headwinds, and in 60 kph tail winds as there is no effective wind resistance at all other than what results from your own progress over the ground.
Wind resistance (drag) is calculated against the speed of the wind, not the speed over ground. Hence why planes go slower over the ground in a headwind, and faster with a tail wind, yet they travel through the air at the same speed.
This is also the reason that wind tunnels work. If you calculated drag as speed over ground, wind tunnels would be absolutely useless.
To put it another way: Wind resistance (drag) is a measure of how much resistance there is for going through the air at a given speed.
power to overcome the drag at 10 mph. 40 mph requires (64) times the power as 10 mph. For ground vehicles add in some for rolling resistance.
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Yes, becaise you're both doing it wrong. Wind resistance is drag through the air, not speed over ground. The only place the two are the same if there is no wind whatsoever.
Ever notice that it is easier to ride fast with a tail wind than it is with a headwind? The reason for that is that your speed through the air changes, and therefore the drag.
Ever notice that it is easier to ride fast with a tail wind than it is with a headwind? The reason for that is that your speed through the air changes, and therefore the drag.
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A commonly known constant in aviation is that the aerodynamic drag increases by the cube of the speed. ie 20mph requires eight ( 8 ) times the
power to overcome the drag at 10 mph. 40 mph requires (64) times the power as 10 mph. For ground vehicles add in some for rolling resistance.
power to overcome the drag at 10 mph. 40 mph requires (64) times the power as 10 mph. For ground vehicles add in some for rolling resistance.
Last edited by CargoDane; 10-30-20 at 12:07 PM.
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Or, more logically, do what I said.
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Unless you are a professional or amateur racer..... Ride as heavy a bike as you can ! You will get a better workout, You will save money as they are cheaper and more rugged requiring less repair, and your crashes will occur at slower speeds. You go out and ride for two hours and come home. Do you really care if you went 30 instead of 40 miles ? The idea was to burn 2500 calories either way.. In addition, everyone who passes you will respect you more than they would if you could blast past them on a whim. All the bike manufacturers spend millions of dollars to brainwash you into thinking you aint cool if you dont have the latest lightest bleeding edge technology. Are you going to let yourself fall victim to that ? Me personally- I like to ride my sixty pound tandem solo. It rides much like a limousine and has many other benefits. Twice I have been able to given broken down motorists a ride back to town.
riding a bike should be a joy, I don't get joy out of riding something heavy and ill suited for the riding your are doing.
Life is too short to not to ride your best lightest bike as much as you can
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Life is too short not to ride the best bike you have, as much as you can
(looking for Torpado Super light frame/fork or for Raleigh International frame fork 58cm)
Life is too short not to ride the best bike you have, as much as you can
(looking for Torpado Super light frame/fork or for Raleigh International frame fork 58cm)
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Yes, becaise you're both doing it wrong. Wind resistance is drag through the air, not speed over ground. The only place the two are the same if there is no wind whatsoever.
Ever notice that it is easier to ride fast with a tail wind than it is with a headwind? The reason for that is that your speed through the air changes, and therefore the drag.
Ever notice that it is easier to ride fast with a tail wind than it is with a headwind? The reason for that is that your speed through the air changes, and therefore the drag.
Otto
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As noted in the equation I gave, drag (the product of k, A and v**2) is proportional to the square of relative wind-speed, v, while power is the product of the inline portion of that drag and rider speed, s. Required power only increases as the cube of speed in still air.
Otto
Otto
And you're wrong when you say "Required power only increases as the cube of speed in still air.". No, that is only the point where the ground speed matches the speed through the air. I can't believe I have to say this again: Your wind resistance is the resistance you meet by going through the air. It doesn't matter if it's you who is moving at hurricane speeds or it's the wind moving at hurricane speeds in relation to the ground.
Seriously, planes would have really weird stalling limits, you wouldn't be able to tell the difference on a bike with headwinds or tailwinds, nor would wind tunnels work at all.
In a headwind, your speed through the air increases even though your ground speed doesn't. You said he was "correct" in his calculation, but he is only right with his calculation if there is no wind. Required power to move through the air doesn't change due to your ground speed. It changes the same whether the air is moving faster or you are moving faster through it. Again: Notice a difference while riding 25 mph in a headwind vs. a tailwind? Force required in the headwind will be higher than in still air, and much higher than with the tailwind.
Last edited by CargoDane; 10-30-20 at 01:19 PM. Reason: move the air > move through the air
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2) Yes, in still air.
3) You two should know better by now as it has been explained ad nauseam to you.
4) If the formulas held true that ground speed is the same as wind speed - even with wind around, you wouldn't be able to tell a difference with your bike between going in still air, in tailwinds, or in headwinds - even at hurricane force winds.
And, finally:
5) It has reached a point where I'm beginning to suspect you're doing this on purpose, that you're actively pretending to not understand things in order to troll. No one can continue to stay, let's call it "that unknowing" unless it's on purpose.
Last edited by CargoDane; 10-30-20 at 08:16 PM.
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With airplanes, the power required to overcome aerodynamic drag simply scales with the cube of air speed.
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In other words, it's proportional to the square of the air speed multiplied by ground speed. In the absence of wind, the equations reduce to a simpler form, and the power scales with the cube of ground speed.
With airplanes, the power required to overcome aerodynamic drag simply scales with the cube of air speed.
In other words, you're arguing that lift and drag forces changes - as well as stalling speeds) for the wings AND propeller/s if the wheels are touching the ground vs. not touching the ground. Let's hope they're not dragging a piece of string on the ground, because suddenly they have to deal with forces multiplied by the ground speed.
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Again, you're wrong. Go do some calculations for me to check. Other than rolling resistance, it doesn't matter if the wind is moving or you are moving. Again: Frigging windtunnels, headwinds, and tailwinds.
LOL, even with that knowledge, you still argue that aerodynamic drag in the air is somehow different than aerodynamic drag if some part is touching the ground (by many, many factors since you want the aerodynamic drag to be multiplied with the ground speed).
In other words, you're arguing that lift and drag forces changes - as well as stalling speeds) for the wings AND propeller/s if the wheels are touching the ground vs. not touching the ground. Let's hope they're not dragging a piece of string on the ground, because suddenly they have to deal with forces multiplied by the ground speed.
You can check any number of references or textbooks; they will not support your argument.
Last edited by tomato coupe; 10-30-20 at 09:36 PM.
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No, you haven't done the calculations. Of course you won't get the same results is you afterwards multiply the speed through the air with the ground speed. Seriously, I'm done with lies.
Especially the part where you argue the drag changes because wheels are the propulsion rather than, say, gravity or a propeller. If dragged changed due to whether the propulsion is in the air, on the ground or in the water for that matter.
So, in other words, drag will change immensely the moment you stop pedaling, the moment you jump over something, or the moment you let gravity take over.
It's air drag. It's a force to do with air. It doesn't matter if the vehicle in question is standing still and the wind moves, or the vehicle moves and the wind doesn't (or both). It's the movement through the air, and with boats, although those also have air drag, it's mostly about drag in the water. It doesn't matter if the propulsion are sails or a propeller.
Show me those "reference books" which supports your claim that it matters what actually propels you. Those same books would also say that drag cannot be calculated in wind tunnels and that headwind and tailwind produces much less drag when you use, say, a tow rope rather than the wheels, or a rocket engine rather than wheels.
You're not even sure about what the consequences of what you're saying have. WHY would what type of propulsion used to push against the force of the effective wind matter. No, I don't need another unsubstantiated claim. I am asking why. Any propulsion system produces a force to counteract other forces to move.
If what you said were true, then the moment one starting applying force while riding (let's say downhill, but could be anything), THEN suddenly the forces to overcome are somehow suddenly multiplied.
The opposite is also true: If you pedalled yourself up to speed downhill and then suddenly stopped pedalling, then suddenly the drag you're trying to overcome would then be divided by your ground speed, meaning you would actually accelerate as the forces you are trying to overcome are suddenly decimated.
No frigging textbooks supports any such claims.
Another thing I want you to explain according to your "understanding" of this: How does tailwinds and headwinds work? How does wind tunnels? This is beyond ridiculous.
Especially the part where you argue the drag changes because wheels are the propulsion rather than, say, gravity or a propeller. If dragged changed due to whether the propulsion is in the air, on the ground or in the water for that matter.
So, in other words, drag will change immensely the moment you stop pedaling, the moment you jump over something, or the moment you let gravity take over.
It's air drag. It's a force to do with air. It doesn't matter if the vehicle in question is standing still and the wind moves, or the vehicle moves and the wind doesn't (or both). It's the movement through the air, and with boats, although those also have air drag, it's mostly about drag in the water. It doesn't matter if the propulsion are sails or a propeller.
Show me those "reference books" which supports your claim that it matters what actually propels you. Those same books would also say that drag cannot be calculated in wind tunnels and that headwind and tailwind produces much less drag when you use, say, a tow rope rather than the wheels, or a rocket engine rather than wheels.
You're not even sure about what the consequences of what you're saying have. WHY would what type of propulsion used to push against the force of the effective wind matter. No, I don't need another unsubstantiated claim. I am asking why. Any propulsion system produces a force to counteract other forces to move.
If what you said were true, then the moment one starting applying force while riding (let's say downhill, but could be anything), THEN suddenly the forces to overcome are somehow suddenly multiplied.
The opposite is also true: If you pedalled yourself up to speed downhill and then suddenly stopped pedalling, then suddenly the drag you're trying to overcome would then be divided by your ground speed, meaning you would actually accelerate as the forces you are trying to overcome are suddenly decimated.
No frigging textbooks supports any such claims.
Another thing I want you to explain according to your "understanding" of this: How does tailwinds and headwinds work? How does wind tunnels? This is beyond ridiculous.
Last edited by CargoDane; 10-30-20 at 10:01 PM.
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Here you go, just substitute bicycle for car:
Under the assumption that the fluid is not moving relative to the currently used reference system, the power required to overcome the aerodynamic drag is given by:
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome aerodynamic drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW).[17] With a doubling of speed the drag (force) quadruples per the formula. Exerting 4 times the force over a fixed distance produces 4 times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, 4 times the work done in half the time requires 8 times the power.
When the fluid is moving relative to the reference system (e.g. a car driving into headwind) the power required to overcome the aerodynamic drag is given by:
Where is the wind speed and is the object speed (both relative to ground).
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome aerodynamic drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW).[17] With a doubling of speed the drag (force) quadruples per the formula. Exerting 4 times the force over a fixed distance produces 4 times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, 4 times the work done in half the time requires 8 times the power.
When the fluid is moving relative to the reference system (e.g. a car driving into headwind) the power required to overcome the aerodynamic drag is given by:
Where is the wind speed and is the object speed (both relative to ground).
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